Question

Tìm x,y biết:

\((x - 11 + y)^2 + (x - 4 - y)^2 = 0\)

Answer 1

Với mọi giá trị của x;y ta có:

\(\left\{{}\begin{matrix}\left(x-11+y\right)^2\ge0\\\left(x-4-y\right)^2\ge0\end{matrix}\right.\Rightarrow\left(x-11+y\right)^2+\left(x-4-y\right)^2\ge0\)

Để \(\left(x-11+y\right)^2+\left(x-4-y\right)^2=0\) thì:

\(\left\{{}\begin{matrix}\left(x-11+y\right)^2=0\\\left(x-4-y\right)^2=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x+y=11\\x-y=4\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}x+y+x-y=11+4\\x+y-x+y=11-4\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}2x=15\\2y=7\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}x=7,5\\y=3,5\end{matrix}\right.\)

Chúc bạn học tốt!!!

Answer 2

Ta có:\(\left(x-11+y\right)^2+\left(x-4-y\right)^2=0\)

mà \(x-11+y\ge0\forall x\) và \(x-4-y\ge0\forall x\)

\(\Rightarrow \begin{cases} x-11+y=0\\ x-4-y=0 \end{cases}\Rightarrow \begin{cases} x+y=11\\ x-y=4 \end{cases}\Rightarrow \begin{cases} x=\dfrac{15}{2}\\ y=\dfrac{7}{2} \end{cases}\)

Vậy \(x=\dfrac{15}{2};y=\dfrac{7}{2}\).

Answer 3

\(\left(x-11+y\right)^2+\left(x-4-y\right)^2=0\)

\(\left\{{}\begin{matrix}\left(x-11+y\right)^2\ge0\\\left(x-4-y\right)^2\ge0\end{matrix}\right.\)

\(\Rightarrow\left(x-11+y\right)^2+\left(x-4-y\right)^2\ge0\)

Dấu "=" xảy ra khi:

\(\left\{{}\begin{matrix}\left(x-11+y\right)^2=0\\\left(x-4-y\right)^2=0\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}x-11+y=0\Rightarrow x+y=11\\x-4-y=0\Rightarrow x-y=4\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}x+y+x-y=11+4\Rightarrow2x=15\Rightarrow x=\dfrac{15}{2}\\y=11-\dfrac{15}{2}=\dfrac{7}{2}\end{matrix}\right.\)

Answer 4

Ta có:

\(\left(x-11+y\right)^2\ge0\forall x,y\\ \left(x-4-y\right)^2\ge0\forall x,y\\ \Rightarrow\left(x-11+y\right)^2+\left(x-4-y\right)^2\ge0\forall x,y\)

Để

\(\left(x-11+y\right)^2+\left(x-4-y\right)^2=0\Rightarrow\left\{{}\begin{matrix}\left(x-11+y\right)^2=0\\\left(x-4-y\right)^2=0\end{matrix}\right.\\ \Rightarrow\left\{{}\begin{matrix}x-11+y=0\\x-4-y=0\end{matrix}\right.\\ \Rightarrow\left\{{}\begin{matrix}2x-15=0\\x-y=4\end{matrix}\right.\\ \Rightarrow\left\{{}\begin{matrix}x=7,5\\x-y=4\end{matrix}\right.\\ \Rightarrow\left\{{}\begin{matrix}x=7,5\\y=3,5\end{matrix}\right.\)

Vậy \(x=7,5;y=3,5\)